In the stellar proton-proton chain reaction, deuterium fuses with a proton to form Helium-3, but deuterium does not fuse directly with another deuterium nucleus. Thus, reaction (2) is not part of this cycle.

According to Einstein's photoelectric equation, \(K_{\text{max}} = h\nu - \phi\). The maximum kinetic energy depends on the frequency/wavelength of the incident light and the work function, but is independent of the intensity of the light.

Second excited state corresponds to \(n = 3\). The energy is \(E_3 = -\frac{13.6}{3^2} = -1.51\text{ eV}\). The ionization energy required is \(E_{\text{ion}} = 0 - E_3 = 1.51\text{ eV}\).

The de Broglie wavelength for an electron accelerated through a potential \(V\) is given by \(\lambda = \frac{1.227}{\sqrt{V}}\text{ nm}\). Substituting \(V = 81\text{ V}\), we get \(\lambda = \frac{1.227}{9}\text{ nm} \approx 0.136\text{ nm}\).

The Lyman series transitions terminate at the ground state (\(n = 1\)). The photon energies emitted in these transitions correspond to the ultraviolet region of the electromagnetic spectrum.

The shortest wavelength in a series corresponds to transition from \(n_2 = \infty\) to \(n_1\). For Lyman series, \(\lambda_L = \frac{1}{R}\). For Balmer series, \(\lambda_B = \frac{4}{R}\). Therefore, the ratio \(\frac{\lambda_L}{\lambda_B} = \frac{1}{4}\).

Intensity of light from a point source is inversely proportional to the square of the distance, \(I \propto \frac{1}{d^2}\). Since the number of photoelectrons emitted per second is proportional to intensity, doubling the distance reduces the emission to one-fourth.

Assertion is true because binding energy per nucleon is the direct measure of nuclear stability. Reason is false because stability does not simply increase with nucleon number; heavy nuclei with a very large number of nucleons become unstable.

Initial BE \(= 240 \times 7.6 = 1824\) MeV. Final BE \(= 2 \times (120 \times 8.5) = 2040\) MeV. Total gain in BE \(= 2040 - 1824 = 216\) MeV.